The function #f(x)=cosx# has limited range between #-1# and #1#. Further its value is #0# at #(2n+1)pi/2#, where #n# is an integer and its value is #1# at #x=2npi# and is #-1# at #(2n+1)pi#. The urve appears as follows
graph{cosx [-10, 10, -5, 5]}
But we have to draw the graph of #f(x)=cos(x+pi/2)#. As cosine of any #x# is always between #-1# and #1#, value of #cos(x+pi/2)# will also be between #-1# and #1#.
The only difference that happens is that instead of #x=0#, as we have #f(x)=cos(x+pi/2)# it is at #-pi/2# that #f(x)=1#. In fact, the entire graph of #cosx# will shift by #pi/2# towards left.
This is called as phase shift.
and #f(x)# will be #0# at #x=npi#, will be #1# at #x=(4n-1)pi/2# and will be #-1# at #x=(4n+1)pi/2#.
The graph appears as shown below.
graph{cos(x+pi/2) [-10, 10, -5, 5]}
